# Charged massive particle reacting to EM wave

1. May 5, 2015

### RawrSpoon

1. The problem statement, all variables and given/known data
Consider a particle of charge q and mass m, free to move in the xy plane in response to an electromagnetic wave propagating in the z direction (might as well set δ to zero)

a) Ignoring the magnetic force, find the velocity of the particle, as a function of time. (Assume the average velocity is zero).

b) Now calculate the resulting magnetic force on the particle

c) Show that the (time) average magnetic force is zero.

The problem with this naive model for the pressure of light is that the velocity is 90 degrees out of phase with the fields. For energy to be absorbed, there's got to be some resistance to the motion of the charges. Suppose we include a force of the form -γmv, for some damping constant γ.

d) Repeat part (a) (ignore the exponentially damped transient). Repeat part (b), and find the average magnetic force on the particle.

2. Relevant equations
$$E(zt) = E_0 cos(k z-\omega t + \delta)\hat{x}$$
$$B(zt) =\frac{1}{c} E_0 cos(k z-\omega t + \delta)\hat{z}$$

3. The attempt at a solution
So I got a,b, and c I'm pretty sure. A quick look over my work by you guys wouldn't be hurtful, I don't think. Thing is, I can't find the solutions manual for the fourth edition of Griffiths's "Introduction to Electrodynamics" anywhere, and this seems to be a new problem.

a) I figured that since we're finding the velocity of this while ignoring the magnetic field, we can find the velocity simply through
$$\int \! F \, \mathrm{d}t=p=mv$$
That
$$v=\frac{\int \! F \, \mathrm{d}t}{m}$$
As such, I got
$$v(t)=-\frac{qE_0}{\omega m}sin(kz-\omega t)\hat{x}$$

b) With this velocity, I remembered that
$$F=qv \times B$$
So with this knowledge, I found that
$$F=\frac{q^2E_0^2}{c \omega m}sin(kz-\omega t)cos(kz-\omega t)\hat{y}$$

c) The average of a function is
$$\frac{1}{T}\int_0^T \! f(t) \, \mathrm{d}t$$

So I found the average of the force over a complete cycle. Because we're in a complete cycle, I realized I had to use
$$\omega=\frac{2\pi}{T}$$
Upon doing the substitution and the integral, I proved the answer was 0. Simple enough.

Now, the part that's really killing me is d. I figure I solve it the same way except the force is replaced with
$$F=qE-\gamma mv$$ However, this isn't proving to be fruitful, as I'm not really sure how to deal with the retarding force. At first I tried the same approach as in part a, with v being replaced as dx/dt which gave me
$$\int \gamma m \mathrm{d}x$$ but that didn't make sense to me, since it'd mean that the retarding force would be based on the position? That doesn't really sit right with me.

I'm open to any suggestions as to where to head to next, and whether or not I'm doing this correctly at all! Thank you so much in advance.

Last edited by a moderator: May 7, 2015
2. May 6, 2015

### Staff: Mentor

It will, indirectly, as velocity depends on position (which you don't know yet).

$F=qE-\gamma mv$ leads to a differential equation. Did you try solving this?

3. May 6, 2015

### RawrSpoon

Funnily enough I JUST tried it about 30ish minutes ago. I think I did it wrong though, it's kind of a large equation and, to be fair, I let Mathematica solve it for me. Anyway, I tried the following
$$m\frac{dv}{dt}+\gamma mv=qE_0cos(kz-\omega t)$$
which led me to
$$v(t)=\frac{qE_0}{m(\gamma ^2+\omega ^2)}(\gamma cos(kz-\omega t)-\omega sin(kz-\omega t))\hat{x}+C_1 e^{-\gamma t} \hat{x}$$
Which looks incredibly complicated. I put in v(0)=0 and got
$$v(t)=\frac{qE_0}{m(\gamma ^2+\omega ^2)}\bigg( \Big( \gamma cos(kz-\omega t)-\omega sin(kz-\omega t) \Big)-e^{-\gamma t}\big(\gamma cos(kz) - \omega sin(kz)\big)\bigg) \hat{x}$$

Seems really complicated. Am I on the right track or should the answer be simpler?

Last edited: May 6, 2015
4. May 6, 2015

### Staff: Mentor

The answer looks great, but don't use v(0)=0.
You can neglect the exponential term (see problem statement), then you get an oscillation that is shifted in phase with respect to the original oscillation. That's exactly what you need to get radiation pressure.

5. May 6, 2015

### RawrSpoon

If I'm understanding you correctly, that means that C1 is 0, right? At least for these purposes.

As such I get that
$$F_{mag}=\frac{q^2E_0^2}{mc(\gamma^2+\omega^2)}\Big(\gamma cos^2(kz-\omega t)-\omega sin(kz-\omega t)cos(kz-\omega t)\Big)\hat{z}$$

Which would give me an average of
$$\frac{q^2E_0^2\gamma}{2mc(\gamma^2+\omega^2)}\hat{z}$$

I think I did it right, the solutions don't have striking errors I don't think?

I can't seem to edit the original post, but the y hat in part b should be a z hat, just a minor correction.

Last edited: May 6, 2015
6. May 7, 2015

### Staff: Mentor

Looks right, the units match and it makes sense in terms of physics. I didn't check the prefactor.

I fixed it.